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If \(x\) is a positive integer, is \(y^2 (x^3 - x + 1) \gt 75?\)

We need to know values of two variables to answer the question:

The value of \(x\)

The value of \(y\)

Statement (1) by itself is insufficient. We only know that \(y \gt 1\).

Statement (2) by itself is insufficient. We know the value of \(x\). Knowing that \(x\) is a positive integer, we conclude that \(x-1 = 4\), thus, \(x = 5\). However, we don't know the value of \(y\).

Statements (1) and (2) combined are sufficient. If \(x = 5\), then \(5^3 - 5 + 1 = 121\). If \(y \gt 1\), then the statement will always be greater than 75.

We need to know values of two variables to answer the question:

The value of \(x\)

The value of \(y\)

Statement (1) by itself is insufficient. We only know that \(y \gt 1\).

Statement (2) by itself is insufficient. We know the value of \(x\). Knowing that \(x\) is a positive integer, we conclude that \(x-1 = 4\), thus, \(x = 5\). However, we don't know the value of \(y\).

Statements (1) and (2) combined are sufficient. If \(x = 5\), then \(x = 5^3 - 5 + 1 = 121\). If \(y \gt 1\), then the statement will always be greater than 75.

Answer: C

I'm not sure I fully understand this. If \(x = 5\) then, then \(x = 5^3 - 5 + 1 = 121\), as such, \(y^2 > 75/121\) which is always true, whether \(y = 1\) or \(y = -1\). That should be the second statement of x=5 is sufficient.

We need to know values of two variables to answer the question:

The value of \(x\)

The value of \(y\)

Statement (1) by itself is insufficient. We only know that \(y \gt 1\).

Statement (2) by itself is insufficient. We know the value of \(x\). Knowing that \(x\) is a positive integer, we conclude that \(x-1 = 4\), thus, \(x = 5\). However, we don't know the value of \(y\).

Statements (1) and (2) combined are sufficient. If \(x = 5\), then \(x = 5^3 - 5 + 1 = 121\). If \(y \gt 1\), then the statement will always be greater than 75.

Answer: C

I'm not sure I fully understand this. If \(x = 5\) then, then \(x = 5^3 - 5 + 1 = 121\), as such, \(y^2 > 75/121\) which is always true, whether \(y = 1\) or \(y = -1\). That should be the second statement of x=5 is sufficient.

What if y=0? Or y=1/2? Or y=-1/2? There are infinitely many values of y, for which y^2>75/121 is not true.
_________________

I think this is a high-quality question and I don't agree with the explanation. As per Statement 2 - X may have 2 possible solutions 5 and -3 so even if we analyse both statement together than also the whole value might not be greater than 75

I think this is a high-quality question and I don't agree with the explanation. As per Statement 2 - X may have 2 possible solutions 5 and -3 so even if we analyse both statement together than also the whole value might not be greater than 75

Please read carefully: If x is a positive integer, is ...
_________________

For this question, since the question asks whether or not y>0 and through these two statements we get (y^2)>75/121 which means that the solution set for y is (-infinity to -(75/121)^(1/2)) U ((75/121)^(1/2) to infinity). Am I missing something here?

For this question, since the question asks whether or not y>0 and through these two statements we get (y^2)>75/121 which means that the solution set for y is (-infinity to -(75/121)^(1/2)) U ((75/121)^(1/2) to infinity). Am I missing something here?

Not sure I understand what you mean...

The question asks whether \(y^2 (x^3 - x + 1) \gt 75\), not whether y > 0. (2) says that y > 0. From (1) the question becomes whether \(y^2 \gt 75/121\) and since from (2) we know that y > 0, then when considering the statements together we can answer that YES \(y^2 \gt 75/121\).
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(x-1)^2=16 x^2-2x+1=16 x^2-2x=15 x(x-2)=15 So x=0 or x=17

Where did I go wrong?

You can equate multiples to 0, when the whole product equals 0. For example, if it were x(x - 2) = 0, then yes, x = 0 or x = 2. Notice that if x = 0, then x(x - 2) = 15, does not hold. Also, it's not clear how you got x = 17, there. Again, if x = 17, then x(x - 2) = 15, does not hold.
_________________

1 clarification needed.. in statement 2-> we get x=5 y^2 * 121>75 to be proven? y^2 > or = to zero even if we take y as 1/2 then y^2 =1/4 1/4 *121=30.25 <75 ---we get answer as NO

Put y=0 then y^2=0 0*121=0 <75 ---we get answer as NO

Put y=-12/11 then y^2=144/121 144/121 *121=144 >75 ---we get answer as YES

if y= integer then definitely y^2 * 121>75

Please let me know what am I lacking in understanding?Acc. to me statement2 is sufficient.

1 clarification needed.. in statement 2-> we get x=5 y^2 * 121>75 to be proven? y^2 > or = to zero even if we take y as 1/2 then y^2 =1/4 1/4 *121=30.25 <75 ---we get answer as NO

Put y=0 then y^2=0 0*121=0 <75 ---we get answer as NO

Put y=-12/11 then y^2=144/121 144/121 *121=144 >75 ---we get answer as YES

if y= integer then definitely y^2 * 121>75

Please let me know what am I lacking in understanding?Acc. to me statement2 is sufficient.

Thanks, Stuti

For some values of y, the answer is YES, for example if y = 100. For some values of y, the answer is NO, for example if y = 0. Two answers, hence insufficient.
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What do you guys think about the variable approach from Math Revolution? I think that in some cases it is better to use this approach, but in most cases - not. And if you are aiming for, at least, Q50, variable approach should not be your friend